Time evolution in quantum mechanics

We know that an operator A in quantum mechanics has time evolution given by Heisenberg equation:

$$\frac{i}{\hbar}[H,A]+\frac{\partial A}{\partial t}=\frac{d A}{d t}$$

Can we derive from this that

$$A(t)=e^{\frac{i}{\hbar}Ht}A(0)e^{-\frac{i}{\hbar}Ht} \qquad ?$$

L.M.: I added $i/\hbar$ in front of $[H,A]$.

Answer

We have to consider the operator, that doesn't explicitly depends on time.

$\frac{\partial A}{\partial t} = 0$

Let's apply commutator formula recursively:

$\frac{d^2 A}{d t^2} = \left(\frac{i}{\hbar}\right)^2[H,[H,A]]$

$\frac{d^3 A}{d t^3} = \left(\frac{i}{\hbar}\right)^3[H,[H,[H,A]]]$

e.t.c.

Then we combine those derivatives in a series for $A(t)$

$A(t)=A(0)+\frac{d A}{d t}t+\frac{1}{2!}\frac{d^2 A}{d t^2}t^2+\frac{1}{3!}\frac{d^3 A}{d t^3}t^3+...$

$A(t)=A(0)+\frac{i}{\hbar}[H,A]t+\frac{1}{2!}\left(\frac{i}{\hbar}\right)^2[H,[H,A]]t^2+\frac{1}{3!}\left(\frac{i}{\hbar}\right)^3[H,[H,[H,A]]]t^3+...$

And then you use this formula to arrive at the result:

$e^XYe^{-X} = Y+\frac{1}{1!}[X,Y]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+...$